a gas has a volume of 1.75 L at -23 degrees C and 150.kPa What temp would the gas occupy 1.30 L at 210.kPa?
It's 260 K
To answer this question, we can use the combined gas law equation, which relates the initial and final conditions of pressure, volume, and temperature of a gas.
The combined gas law equation is:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Here, P1, V1, and T1 represent the initial conditions, while P2, V2, and T2 represent the final conditions.
Given:
Initial conditions:
P1 = 150 kPa
V1 = 1.75 L
T1 = -23 degrees C (We need to convert it to Kelvin for the equation)
To convert Celsius to Kelvin, we add 273.15 to the Celsius value:
T1 = -23 + 273.15 = 250.15 K
Final conditions:
V2 = 1.30 L
P2 = 210 kPa
T2 = unknown (We need to solve for it)
Now, let's substitute these values into the equation and solve for T2:
(150 kPa * 1.75 L) / (250.15 K) = (210 kPa * 1.30 L) / (T2)
Cross-multiplying:
150 kPa * 1.75 L * T2 = 210 kPa * 1.30 L * 250.15 K
Calculating:
(150 kPa * 1.75 L * T2) / (210 kPa * 1.30 L) = 250.15 K
T2 = (210 kPa * 1.30 L * 250.15 K) / (150 kPa * 1.75 L)
Simplifying:
T2 = (68264.125 kPa∙L∙K) / (183.75 kPa∙L)
T2 ≈ 372.05 K
Therefore, the gas would occupy 1.30 L at 210 kPa at a temperature of approximately 372.05 Kelvin.